Problem 18 - Calculus

Question

Find the derivatives of the functions \(9-18:\) \(\ln (\operatorname{sech} x+\tanh x)\)

Step-by-Step Solution

Answer

The derivative is \[ \frac{dy}{dx} = \frac{-\operatorname{sech} x \cdot \tanh x + \operatorname{sech}^2 x}{\operatorname{sech} x + \tanh x}. \]

Step 1: Identify the Function

The function provided is \( y = \ln (\operatorname{sech} x + \tanh x) \). Our goal is to find its derivative, \( \frac{dy}{dx} \).

Step 2: Apply the Chain Rule

To differentiate \( y = \ln (u) \), where \( u = \operatorname{sech} x + \tanh x \), we use the chain rule. This gives us:\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}. \]

Step 3: Differentiate the Inner Function

Find the derivative of the inner function \( u = \operatorname{sech} x + \tanh x \). The derivative of \( \operatorname{sech} x \) is \( -\operatorname{sech} x \cdot \tanh x \), and the derivative of \( \tanh x \) is \( \operatorname{sech}^2 x \). Therefore, \[ \frac{du}{dx} = -\operatorname{sech} x \cdot \tanh x + \operatorname{sech}^2 x. \]

Step 4: Substitute and Simplify

Substitute \( u \) and \( \frac{du}{dx} \) back into the formula from Step 2:\[ \frac{dy}{dx} = \frac{1}{\operatorname{sech} x + \tanh x} \left(-\operatorname{sech} x \cdot \tanh x + \operatorname{sech}^2 x\right). \] Simplifying this expression can depend on further algebraic manipulation if needed.

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus essential for differentiating composite functions. Believe it or not, it simplifies the process by breaking down the differentiation into smaller, more manageable parts. The mantra to remember is "outer function times derivative of the inner function".

Whenever you have a function nested within another, like \( y = \ln(u) \) with \( u = \operatorname{sech} x + \tanh x \), the chain rule comes to our rescue. It tells us to take the derivative of the outer function first, which is \( \ln(u) \), and multiply it by the derivative of the inner function \( u \).

• Differentiate the outer function, treating the inside as constant: For \( \ln(u) \), it's \( \frac{1}{u}\).
• Multiply by the derivative of the inner function \( u \).

A powerful technique, it allows you to tackle tougher differentiations with relative ease, step by step.

Exploring Hyperbolic Functions

Hyperbolic functions, much like their trigonometric siblings, appear often in calculus, especially when dealing with exponential functions. Here, we encounter \( \operatorname{sech} x \) and \( \tanh x \). Knowing their derivatives is crucial for solving our exercise.

• The hyperbolic secant function is \( \operatorname{sech} x = \frac{1}{\cosh x} \). Its derivative happens to be \(-\operatorname{sech} x \cdot \tanh x \).
• The hyperbolic tangent function, \( \tanh x = \frac{\sinh x}{\cosh x} \), simplifies to a derivative of \( \operatorname{sech}^2 x \).

These derivatives are vital in calculating how \( u \) changes, as seen in the chain rule application. They illustrate how hyperbolic functions shadow their trigonometric counterparts, making them easier to grasp if you're already familiar with trigonometry.

The Role of the Natural Logarithm

The natural logarithm, \( \ln(x) \), is more than a calculator button. It represents the logarithm with a base of \( e \), the mathematical constant approximately equal to 2.718. It's widely used in growth processes, time calculations, and, as in this case, differentiations.

In differentiation, \( \ln(x) \) follows a straightforward rule: \( \frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx} \). This method is the basis of our chain rule example. Its beauty lies in its simplicity, breaking down even complex expressions into something manageable. Just remember, the natural logarithm efficiently handles multiplicative and exponential relationships, making it an asset in the world of differentiation.